I have $X_n \overset{d}{\rightarrow} N(\mu_1, \sigma^2_1)$, and $Y_n \overset{d}{\rightarrow} N(\mu_2, \sigma^2_2)$, and $X_n$ and $Y_n$ are independent, for $n \in \mathbb{N}$.
(Note that it does not imply $X_i$ and $Y_j$ are independent where $i \neq j$)
Can we say as follows, $(X_n, Y_n) \overset{d}{\rightarrow} N_2\left((\mu_1, \mu_2), \begin{bmatrix} \sigma^2_1 & 0 \\ 0 & \sigma^2_2 \end{bmatrix}\right)$?
My intuition is strongly positive about this, but I need a proof for this.
Let $(U,V)\sim N((\mu_1,\mu_2),\operatorname{diag}(\sigma_1^2,\sigma_2^2))$. We claim $(X_n,Y_n)\xrightarrow{d}(U,V)$. Indeed, by Levy's theorem it suffices to check characteristic function $$ \begin{align*} \varphi_{(X_n,Y_n)}(s,t)&= \mathbb{E}[e^{i(s,t)\cdot(X_n,Y_n)}]\\ &=\mathbb{E}[e^{isX_n}]\mathbb{E}[e^{itY_n}]\\ &\to\mathbb{E}[e^{isU}]\mathbb{E}[e^{itV}]\\ &=\mathbb{E}[e^{i(s,t)\cdot(U,V)}]\\ &=\varphi_{(U,V)}(s,t) \end{align*} $$